Please post your work. 1.) Two identical diverging lenses are separated by 15 cm

Question

Please post your work.

1.) Two identical diverging lenses are separated by 15 cm. The focal length of each lens is -8.0 cm. An object is located 4.0 cm to the left of the lens that is on the left. Determine the final image distance relative to the lens on the right.
_____________cm

2.) A slide projector has a converging lens whose focal length is 102.00 mm. (a) How far from the lens must the screen be located if a slide is placed 104 mm from the lens?
m

(b) If the slide measures 25.0 mm × 37.0 mm, what are the dimensions (in mm) of its image?

Explanation / Answer

1)
given f1 = f2 = -8cm

for first lense

object distance,u1 = 4 cm

focal length, f1 = -8cm cm

let v1 is the image distance for the first lense.

1/u1 + 1/v1 = 1/f1

1/v1 = 1/f1 - 1/u1

1/v1 = 1/(-8) - 1/4

v1 = -2.67 cm

object distance for second lense,u2 = -(15+2.67)

= -17.67 cm

let v2 is the image distance for the second lense.

1/u2 + 1/v2 = 1/f2

1/v2 = 1/f2 - 1/u2

1/v2 = 1/(-8) - 1/(17.67)

v2 = -5.5 cm

so image distnace is 5.5 cm to the left of second lense.

2)

a)given f = 102 mm

object distance, u = 104 mm

let v is the image distance.

Apply, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/102 - 1/104

v = 5304 mm

= 5.304 m

b) magnification, m = -v/u

= -5304/104

= -51

|m| = 51

image width = object width*m = 25*51 = 1275 mm

image height = object height*m = 37*51 = 1887 mm

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